Positron in a magnetic field

  • Thread starter elitepro
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  • #1
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Homework Statement


A positron moves in a circular path of radius R due to a uniform magnetic field of strength B applied perpendicular to the plane of the circle. If B is varied, which of the following best represents a graph of the kinetic energy of the positron as a function of B so that the positron maintains the same radius R.

Homework Equations


qvB=mv^2/R

The Attempt at a Solution


KE = 1/2MV^2

Rearranging first equation, RqvB =mv^2, then multiply each side by 2

2RqvB = KE

Here i thought that KE is proportional to B, but v remains a variable that depends on KE. How do create a relationship between B and KE?
 

Answers and Replies

  • #2
TSny
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Try using qvB=mv2/R to get an expression for v.
 
  • #3
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Wow. I don't know how I messed that up. Graph is quadratic. Thanks for your help!
 
  • #4
TSny
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Good work.
 
  • #5
2RqvB = KE

You could also note that v is proportional to KE^(1/2), and then since you don't need to worry about constants such as 1/2m you can just say that 2q KE^1/2 B is proportional to KE, and by dividing KE ^1/2 on both sides you get 2qB is proportional to KE^1/2.
so KE is proportional to B^2 and you get the answer that it is a quadratic formula centered at 0.
 

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